Low pass filter?

Discussion in 'Homework Help' started by Tobias Hildebrandt, Jan 21, 2016.

  1. Tobias Hildebrandt

    Thread Starter Member

    Jan 1, 2016
    44
    0
    Hello,

    U(2)t is supposed to differ from U(1)t by - 45 degrees. Q8 asks for C1 and U2(t). The rest of the information is written down on the circuit.

    So I made to assumptions about this circuit:
    1) R1 and R2 are a Voltage Divider. Since R1 = R2 = 10kOhm. I figured, that we get a voltage of 2.5V as the "new" input signal.

    2) My second assumption is that we deal here with a LOW PASS filter.

    If this is the case, than I could use the following two formulas to solve the problem:

    φ(ω) = -arctan (ωCR)

    -45 = -arctan (ωCR) // Comment *(-1)

    45 = arctan (ωCR) // *tan()

    tan(45) = ωCR // Divided by ωR

    tan(45) / ωR = tan(45) / (2 * π * 1000Hz * 5000Ω) = 32nF = C

    U2(t) = ?

    At Cut off frequency (and that is what we are talking about here, isn't it? (φ = -45°)) I can use U2(t) ≈ U1(t) * √2 = 3.54 V

    I knew the "right" result after I calculated the 32nF for C and it got "confirmed" when I calculated U2(t).

    After I checked the results, it turned out that my assumptions had one tiny problem, they were all bollocks.

    Any idea where I went wrong?

    Tobias

    Capture40.PNG
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,761
    4,800
    At first glance you are not taking into account the output impedance of your new 2.5 V input signal, which is 5 kΩ.

    Don't be so afraid to redraw the circuit you are working with to reflect the simplifications you make.
     
    Tobias Hildebrandt likes this.
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,960
    1,097
    As for the capacitor value
    Fz = 1/(2*pi*RC ) where
    R - is a resistance (Rth) seen by a capacitor. And for your circuit Rth = R1||R2 + R3 = 10kΩ
    So, the capacitor value is C ≈ 0.16/(10kΩ 1kΩ) = 16nF
     
    Tobias Hildebrandt likes this.
  4. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi,

    As Wbahn pointed out, the impedance of the resistor divider has to be taken into account. What you tried to do was replace the resistor voltage divider with an ideal voltage source, and that doesnt work because the resistor divider has impedance of it's own. If you do a Thevenin/Norton equivalent you'll find this out.

    If you were to insert a buffer amplifier with gain of 1 between the divider and the 5k ohm resistor, you would come up with the result you got because then the output of the buffer would approximate an ideal voltage source. But because there is interaction between the resistors and the 5k ohm resistor there will be a loading effect which will change that.

    Another way to look at this is if you replaced the resistors R1 and R2 with 1 ohm resistors (in theory only) you would come up with the approximate result you got. That is because two 1 ohm resistors have very low impedance compared to the 5k ohm. But as the two resistor values rise we loose this relationship so at 10k each the result is C1=16nf and at 100k each the result is 0.32nf, which as you can see are very different values.

    To find the Thevenin/Norton equivalents, you could start by transforming the input voltage source into a current source, and then that simplifies the circuit a little, then convert back to a voltage source, and you'll have the correct equivalents. The voltage equivalent will still work out to 2.5v, but there will also be an equivalent resistance to deal with.

    Starting with the input source and first resistor R1, we get an equivalent current source:
    I1=Vin/R1

    and that means we have a current source I1 in parallel to R1 and because that is in parallel to R2 we now have the two resistors R1 and R2 in parallel, which makes it easier to calculate the total resistance there:
    Rp=R1*R2/(R1+R2)

    and so now the equivalent voltage source is:
    Vx=I1*Rp

    and that is in series with a resistance Rp, and because now Rp is in series with R3, we have an equivalent resistance of:
    Rx=Rp+R3

    So the voltage source has been transformed into Vx and the series resistance into Rx, just one resistor and one voltage source.

    All this was possible because we used the two rules:
    1. A voltage source in series with a resistance is equivalent to a current source in parallel with a resistance.
    2. A current source in parallel with a resistance is equivalent to a voltage source in series with a resistance.

    By using #1 and #2 above we can perform a series of transformations that take us from input to output and eventually end up with either a single voltage source and a single impedance or a single current source and single impedance.
    You would do very well to do more examples of using those two rules of equivalency because they can help simplify circuits so much.
     
    Tobias Hildebrandt likes this.
  5. Tobias Hildebrandt

    Thread Starter Member

    Jan 1, 2016
    44
    0
    Fair enough! So, I cant just ignore the resistors from the voltage divider. :) If use R(Th), than you get the 16nF. Thank you!

    What about U2(t)?
     
  6. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi,

    Do you know how to do complex math?
    Example: 3.2+8j
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,960
    1,097
    But WBahn already explain this to you. Since Rth = 10kΩ and Vth = 2.5V ----> U2(t) = 2.5V*√2 = 2.5V*0.707 = 1.767V.
     
    Tobias Hildebrandt likes this.
  8. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi there Jony,

    I assumed he wanted the full time response u(t) which includes a phase shift, or alternately just the amplitude and phase shift. This would require more than just the amplitude |V|. I could be wrong of course :)
     
  9. WBahn

    Moderator

    Mar 31, 2012
    17,761
    4,800
    It's a multiple choice question (easy to miss since the problem is in German) and the options only give amplitudes.
     
  10. Tobias Hildebrandt

    Thread Starter Member

    Jan 1, 2016
    44
    0
    @WBahn. Yes, Iam very thankful that you are patient with me and I am sorry about the German questions. I post them so that you can see the circuit, I thought that might be helpful to the people that wanted to help me.

    In regards to u(2)t, I am not sure what I was thinking. My formula tells me, that i am supposed to divide U1(t) by sqrt(2) and somehow, I multiplied U1(t) by sqrt(2), which is wrong. I really need to stop making such stupid mistakes.

    Thanks again to everyone!!! You are great bunch of very helpful people! Thank you!
     
  11. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi,

    Yes i always like to do more calculations than needed :)
     
  12. WBahn

    Moderator

    Mar 31, 2012
    17,761
    4,800
    No need to apologize there -- schematics are not only helpful, they are usually crucial.

    Asking sanity-check questions will help with that (and you, me, and everyone else will always be making stupid mistakes).
     
Loading...